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SCH 4U EXAM REVIEW
UNIT 1: QUANTUM; STRUCTURE AND PROPERTIES
Chapter 3- Quantum
1. Periodic trends:
2. Orbits vs. Orbitals
orbit
– 2-D path; fixed distance from nucleus; circular or elliptical path;
orbital
– 3-D region in space; variable distance from nucleus; no path and varied
shape of region; 2 electrons per orbital;
- Solved wave functions in Schrodinger’s equation
3. QUANTUM NUMBERS TO DESCRIBE ANY ELECTRON IN AN ATOM
Quantum numbers
Meaning
n
l
ml
ms
For any n value, there are n2 orbitals, n types of orbitals and up to 2n2 number of
electrons
1
4. Quantum number summary table
n
l
# orbitals (n2)
in this sublevel
ms
ml
Max #
electrons
(2n2)
1
2
3
4

In each energy level, there are:___s-orbital;
p-orbitals;
d-orbitals;
The maximum number of electrons in each orbital:
1s: ____, 2s: ____ , 2p: ____ , 3s: ____ , 3p: ____ , 3d: ____
1. Are the following sets of quantum numbers possible? Why or why not?
a. n = 1, l = 1, ml = 0, ms = ½
b. n = 4, l = 0, ml = 2, ms = ½
5. Rule for filling e- in orbitals
A- the ______________ principle:
sequentially filled based on the sub-shell’s energy level
B- Use the sequence established on the periodic table
2
f-orbitals
C- Each orbital will hold a maximum of two electrons that spin in opposite directions –
the ________________________________ principle
D- Electrons must be distributed among orbitals of equal energy so that as many
electrons remain unpaired as possible – ________________rule
6. Energy-level diagram
e.g. F
7. Anomalous e- config:
Element
e- config predicted by Aufbau
principle:
ending in s2d9
Actual e- config:
ending in s1d10
Copper
[Ar]
[Ar]
[Kr]
[Kr]
Silver
- Promoting an e- onto the higher energy s orbital (e.g. 4s) relieves the repulsion
that would otherwise be experience in the compact 3d orbitals.
Why such
- as a result, actual e- config prefers d10
discrepancy? arrangement meaning that all d orbitals are
completely-filled; repulsion is kept to a
minimum
Element
e- config predicted by Aufbau
principle:
ending in s2d4
Actual e- config:
ending in s1d5
Chromium
[Ar]
[Ar]
Why such
discrepancy?
Element/Ions
all d orbitals are half-filled; repulsion is kept to a
minimum
Short-hand e- config
Fe
Fe2+
3
Chapter 4: Chemical Bonding
1. Types of Bonds –Bond continuum
2. Writing Lewis structure
e.g. SO3
How to calculate Formal
Charge on each atom
FC = # valence electrons - # nonbonding electrons – ½ # bonding
e-
PCl5 (doesn’t obey octet rule)
3. VALENCE BOND THEORY

Covalent bonds form orbitals overlap to share e-.
 (SIGMA)
 (PI)
FOUND IN WHAT BONDS
ORBITALS OVERLAPPED
RELATIVE STRENGTH
4
4. HYBRIDIZATION
Show steps to hybridization of CH4
Molecular shape:
5. VSEPR THEORY
Valence-Shell Electron-Pair Repulsion Theory – pairs of electrons in the valence shell
of an atom stay as far apart as possible to minimize repulsion of their negative charges
1.
2.
3.
4.
Draw the Lewis structure for the molecule, including the e- pairs around the central
atom.
Count the total number of bonding pairs and lone pairs of electrons around the
central atom.
Use the table below to predict the shape of the molecule.
# OF e- PAIRS
AROUND
CENTRAL
ATOM
2
3
ORIENTATION
OF
ELECTRON
PAIRS
linear
triangular
planar
NUMBER OF
BONDING
AND LONE
PAIRS
2 BP, 0 LP
3 BP, 0 LP
BOND
ANGLES
SHAPE
EXAMPLE
180°
linear
BeF2
120°
trigonal
planar
BF3
MOLECULAR
GEOMETRY
X–A–X
X
|
A
X
tetrahedral
4 BP, 0 LP
109.5°
tetrahedral
X
X
|
A
CH4
X
4
X
X
tetrahedral
3 BP, 1 LP
< 109.5°
trigonal
pyramidal
NH3
PCl3
A
X
X
X
5
tetrahedral
2 BP, 2 LP
< 109.5°
v-shaped
or bent
H2O
OF2
tetrahedral
1 BP, 3 LP
180°
linear
HCl
5
trigonal
bipyramidal
5 BP, 0 LP
120° &
90°
trigonal
bipyramidal
PCl5
6
octahedral
6 BP, 0 LP
90°
octahedral
SF6
A
X
e.g., Try ClF3
6. INTERMOLECULAR FORCES
6
X
A–X
X
X
|
X
A
| X
X
X
X
|
X
A
X | X
X
7. STRUCTURE AND PROPERTIES OF CRYSTALLINE SOLIDS (AGGREGATES)
CRYSTAL
PARTICLES
PROPERTIES
COND SOLUBILITY
MELTING
UCIN LIQUID/
POINT
TIVITY
SOLUTION
FORCE/
BOND
EXAMPLES
Ionic
medium
ions
(+ and -)
(higher as
ionic
charge
increases,
size of ion
decreases
)
ionic
yes
yes
NaCl,
Na3PO4,
CaF2, MgO,
CuSO4•5H2
O
yes
no
Pb, Fe,
Cu, Al, Ag,
Au
yes
PF3, ICl,
CHCl3, H2O,
NH3, SO3
no
inert gases,
diatomic
elements,
CO2, CCl4,
SF6, BF3
no
diamond,
SiC, AlN,
BeO, SiO2,
CuCl2, Mg2S
Metallic
metallic
cations
(fixed nuclei
with mobile
delocalized
electrons,
allowing for
conduction of
heat and
electricity)
high
(higher as
number of
valence
electrons
increases)
Polar Molecular
low
(higher as
number of
electrons,
polarity of
molecules
, and
presence
of
hydrogen
bonds
increases)
molecules
London,
dipoledipole,
hydrogen
Nonpolar Molecular
molecules
or single
atoms
Covalent Network
Crystal
no
covalent
atoms
(a 3-D
arrangement
of strong
covalent
bonds
between
atoms;
resulting in
hardness
and high
melting
point)
7
very high
(melting
point
increases
as
strength of
covalent
bonds
increase )
no
UNIT 2- ORGANIC CHEMISTRY
ORGANIC COMPOUNDS
Hydrocarbons
Hydrocarbon Derivatives
Aliphatic
Alkanes
C(O)N
Alkenes
_______
_______
Aromatic
Alkynes
Cyclic
_______
Alkyl Halides
Alcohols
Aldehydes
Amides
:
:
:
:
R-X
R-OH
R-CHO
R-
Amines
:
Carboxylic acids:
Ethers
:
Esters
:
Ketones
:
R-N
RCOOH
R-O-R
RC(O)OR
RC(O)R
BASIC NAMING
NUMBER
1
2
3
4
5
6
7
8
9
10
PREFIX FOR NAMING
methethpropbutpenthexheptoctnondec-
ALKYL GROUP NAME
methylethylpropylbutylpentylhexylheptyloctylnonyldecyl-
FUNCTIONAL GROUP
-F
- Cl
- Br
-I
- OH
- NO2
- NH2
PREFIX FOR NAMING
fluoro
chloro
bromo
iodo
hydroxy
nitro
amino
ISOMERS OF ALKYL GROUPS
butyl (or n-butyl)
CH3 – CH2 – CH2 – CH2
CH3 – CH – CH3
SYMBOLS
R, R΄, R˝  alkyl group
X  halogen atom
isobutyl (or i-butyl)
CH2
(O)  oxidizing agent
s-butyl (secondary butyl)
CH3 – CH – CH2 – CH3
(like KMnO4 or Cr2O7 2- in
H2SO4)
CH3
t-butyl (tertiary butyl)
CH3 – C – CH3
PRIORITY FOR NAMING (FROM HIGHEST TO LOWEST)
Highest: Carboxylic acids > Esters > Acid Halides> Amides> Nitriles> Aldehydes>
Ketones> Alcohols> Phenols> Amines> Imines> Alkenes> Alkynes> Alkanes
8
ISOMERS: molecules of the _____________ molecular formula but different
____________________
Alkenes have ___________ isomers. e.g.
FLOW CHART OF ORGANIC REACTIONS
POLYMERS
1. Addition polymers: require ______________________
2.
3. Condensation polymers: require ____________________________________
Final by-product: ____________
9
CHAPTER 5: THERMOCHEMISTRY
thermochemistry – the study of the energy changes that accompany physical or
chemical changes of matter
thermal energy – energy available from a substance as a result of the motion of its
molecules
chemical system – a set of reactants and products under study, usually represented by
a chemical equation
surroundings – all matter around the system that is capable of absorbing or releasing
thermal energy
heat – amount of energy transferred between substances
exothermic – releasing thermal energy as heat flows out of the system; negative molar
enthalpy
endothermic – absorbing thermal energy as heat flows into the system; positive molar
enthalpy
temperature – average kinetic energy of the particles in a sample of matter
open system – one in which both matter and energy can move in or out (e.g., burning
marshmallow)
isolated system – an ideal system in which neither matter nor energy can move in or
out (e.g., ideal calorimeter)
closed system – one in which energy can move in or out, but not matter (e.g., realistic
calorimeter)
calorimetry – the process of measuring energy changes in a chemical system
enthalpy change (ΔH) – the difference in enthalpies of reactants and products during a
change
q = quantity of heat (J) = m c ΔT
m
n M
MOLAR ENTHALPY
molar enthalpy (ΔHx) – the enthalpy change associated with a physical, chemical, or
nuclear change involving one mole of a substance; examples are
below:
TYPE OF MOLAR ENTHALPY
solution (ΔHsol)
combustion (ΔHcomb)
vaporization (ΔHvap)
freezing (ΔHfr)
neutralization (ΔHneut)
formation (ΔHf)
ΔH = n ΔHvap or sol = mcΔT
EXAMPLE OF CHANGE
NaBr(s)  Na+(aq) + Br-(aq)
CH4(g) + 2 O2(g)  CO2(g) + H2O(l)
CH3OH(l)  CH3OH(g)
H2O(l)  H2O(s)
2 NaOH(aq) + H2SO4(aq)  2 Na2SO4(aq) + 2 H2O(l)
C(s) + 2 H2(g) + ½ O2(g)  CH3OH(l)
ΔH = q
10
n
ASSUMPTIONS USED IN CALORIMETRY:
1. No heat is transferred between the calorimeter and the outside environment.
2. Any heat absorbed or released by the calorimeter materials is usually negligible.
3. A dilute aqueous solution is assumed to have a density and specific heat capacity
equal to that of pure water (1.00 g/mL and 4.18 J/g•°C).
LAB: COMBUSTION OF ALCOHOLS





Mass of Fuel Burned:
1.41 g
Mass of Water Heated:
97.00 g
Mass of Water Vapourized: 0.18 g
Mass of Pop Can:
16.07 g
Temperature Change of Water:
22.2 °C

Heat Absorbed by Water:

Heat Absorbed by Can:
q = m c Δt
= (16.07 g) (0.900 J / g · °C) (22.2 °C)
= 321 J

Heat Used to Vapourize Water:
q = m · LV
= (0.18 g) (2268 J / g)
= 4.0 x 102 J

Total Heat Evolved by Fuel: qtotal = (9.00 x 103 J) + (321 J) + (4.0 x 102 J)
= 9.72 x 103 J
= 9.72 kJ

Number of Moles of Fuel:

Molar Heat of Combustion of Fuel: ΔH = qtotal
n
= 9.72 kJ
0.0235 mol
= 414 kJ/mol

% error: 100% –
q = m c Δt
= (97.00 g) (4.18 J / g · °C) (22.2 °C)
= 9.00 x 103 J
n=m
M
= 1.41 g
60.11 g / mol
= 0.0235 mol
experimental enthalpy
actual enthalpy
x 100%
= 100% –
414 kJ x 100%
490 kJ
= 15.5%, therefore unacceptable (>10%)
REPRESENTING ENTHALPY CHANGES
METHOD 1: THERMOCHEMICAL EQUATIONS WITH ENERGY TERMS
e.g., C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l) + 2802.7 kJ
11
METHOD 2: THERMOCHEMICAL EQUATIONS WITH ΔH VALUES
e.g., C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
ΔH = -2802.7 kJ
METHOD 3: MOLAR ENTHALPIES OF REACTION
e.g., ΔHrespiration = -2802.7 kJ/mol glucose
METHOD 4: POTENTIAL ENERGY DIAGRAM
e.g.,
Ep (kJ)
C6H12O6(s) + 6 O2(g)
ΔH = -2802.7 kJ
6 CO2(g) + 6 H2O(l)
Reaction Progress
STANDARD ENTHALPY OF FORMATION
standard enthalpy of formation (ΔH°f) – the quantity of energy associated with the
formation of one mole of a substance from its
elements in standard state; zero for elements in
standard state
HESS’S LAW (ADDITIVITY OF REACTION ENTHALPIES)
Hess’s law – the value of the ΔH for any reaction that can be written in steps equals the
sum of the values of ΔH for each of the individual steps (i.e., ΔHtarget = Σ
ΔHknown)
e.g.
Determine the enthalpy change involved in the formation of two moles nitrogen
monoxide from its elements.
N2(g) + O2(g)  2 NO(g)
ΔH°1 = +34 kJ
ΔH°2 = -56 kJ
(1) ½ N2(g) + O2(g)  NO2(g)
(2) NO(g) + ½ O2(g)  NO2(g)
ΔH°1 = 2(+34) kJ
ΔH°2 = -2(-56) kJ
2 x (1): N2(g) + 2 O2(g)  2 NO2(g)
-2 x (2): 2 NO2(g)  2 NO(g) + O2(g)
ΔH° = +68 kJ + 112 kJ = +180 kJ
N2(g) + O2(g)  2 NO(g)
USING STANDARD ENTHALPIES OF FORMATION TO DETERMINE ΔH
ΔH =
Σ nΔH°f (products)
–
Σ nΔH°f (reactants)
e.g., MULTI-STEP CALCULATION
If 3.20 g of propane burns, what temperature change will be observed if all of the heat from
combustion transfers into 4.0 kg of water?
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
ΔH°f (CO2) = -393.5 kJ/mol
ΔH°f (H20) = -285.8 kJ/mol
12
ΔH°f (C3H8) = -104.7 kJ/mol
ΔH°f (O2) = 0.0 kJ/mol
mpropane = 3.20 g
mH2O = 4.0 kg
cH2O = 4.18 J/(g•°C)
ΔH = Σ nΔH°f (products) – Σ nΔH°f (reactants)
= 3 mol x -393.5 kJ
+
4 mol x -285.8 kJ
–
1 mol x -104.7 kJ
+
5 mol x 0.0
kJ
1 mol
1 mol
1 mol
1
mol
= -2219 kJ
ΔHc (propane) = qwater
n ΔHc
= mcΔT
ΔT
= n ΔHc
mc
mpropane ΔHc
Mpropane mwater c
=
=
(3.20 g) (2219 kJ)
(44.11 g) (4.0 kg) (4.18 J/g•°C)
= 9.6°C
CHAPTER 6: CHEMICAL KINETICS
chemical kinetics – the area of chemistry that deals with rates of reaction
rate of reaction – the speed at which a chemical change occurs, generally expressed
in concentration per unit time, such as mol/(L•s)
rate = Δc
Δt
average rate of reaction – the speed at which a reaction proceeds over a period of
time; determined using slope of a secant (line between two
points)
instantaneous rate of reaction – the speed at which a reaction is proceeding at a
particular point in time; determined using a tangent
METHODS TO MEASURE RATE: pH change; conductivity; volume of gas produced; change
in mass of products; change in colour
RATE LAW EQUATION
r = k [X]m [Y]n
dependence
k = rate constant; valid only for a specific temperature
[X] and [Y] = concentrations of reactants
m and n = order of reaction (describes the initial concentration
of a particular reactant)
overall order of reaction – the sum of the exponents in the rate law equation
e.g., r = k[BrO3(aq)-]1 [HSO3(aq)-]2, therefore overall order is 3 (1 + 2)
13
zeroth-order reaction – the rate does not depend on [A]
e.g., if the initial concentration of A is
doubled, the rate will multiply by 1 (20), and
so will be unchanged
first-order reaction – the rate is directly proportional to [A]
e.g., if the initial concentration of A is doubled,
the rate will multiply by 2 (21)
second-order reaction – the rate is proportional
to the square of [A]
e.g., if the initial
concentration of A is
doubled, the rate will multiply
by 4 (22)
e.g.,
NO(g) + H2(g)  HNO2(g)
EXPERIMENT
NO (MOL/L)
H2 (MOL/L)
1
2
3
4
5
6
0.001
0.002
0.003
0.004
0.004
0.004
0.004
0.004
0.004
0.001
0.002
0.003
a) Write the rate law for the reaction.
r = k [NO(g)]2 [H2(g)]1
INITIAL RATE OF
REACTION (MOL/(L•S)
0.002
0.008
0.018
0.008
0.016
0.024
The units for rate constant, k, are
related to overall order of reaction:
b) Write the overall order of the reaction.
3rd order
c) Calculate k for the reaction (use for experiment’s values).
r = 0.02 mol/L•s
r = k [NO(g)]2 [H2(g)]
[NO] = 0.001 mol/L
(0.02) = k (0.001)2 (0.004)
[H2] = 0.004 mol/L
k = 5 x 106 L2/(mol2•s)
 first order overall, the units are 1/s
or s-1
 second order overall, the units are
L/(mol•s) or L/(mol-1•s-1)
 third order overall, the units are
L2/(mol2•s) or L/(mol-2•s-1)
COLLISION THEORY
CONCEPTS OF THE COLLISION THEORY:
 A chemical system consists of particles (atoms, ions, or molecules) that are in
constant random motion at various speeds. The average kinetic energy of the
particles is proportional to the temperature of the sample.
 A chemical reaction must involve collisions of particles with each other or the walls
of the container.
14



An effective collision is one that has sufficient energy and correct orientation of the
colliding particles so that bonds can be broken and new bonds formed.
Ineffective collisions involve particles that rebound from the collision unchanged.
The rate of a given reaction depends on the frequency of collision and the fraction
of those collisions that are
activated complex
effective.
activation energy – the minimum
increase in
potential energy
of a system
required for
molecules to
react
activation
energy
Ep
products
net potential energy
change, ΔH
reactants
activated complex – an unstable
chemical
species
Reaction
Progress
containing partially broken and partially
formed
bonds representing
the maximum potential energy point in the change; also called the
transition state
reaction mechanism – a series of elementary steps that makes up an overall reaction
elementary step – a step in a reaction mechanism that only involves one-, two-, or
three-particle collisions
rate-determining step – the slowest step in a reaction mechanism
reaction intermediates – molecules formed as short-lived products in reaction
mechanisms
e.g.,
elementary step
rate-determining step
HBr(g) + O2(g)
 HOOBr(g)
HOOBr(g) + HBr(g)
 2 HOBr(g)
2 HOBr(g) + HBr(g)
 H2O(g) + Br2(g) (fast)
(slow)
(fast)
reaction
mechanism
 2 H2O(g) + 2 Br2(g)
4 HBr(g) + O2(g)
reaction intermediate
FACTORS AFFECTING RATE OF REACTION
1. NATURE OF THE REACTANTS:
 each reactant contains a different number of bonds, each with differing bond
strengths, that must be broken for the reaction to proceed
 each reactant has a different threshold energy (minimum kinetic energy required to
convert kinetic energy to activation energy)
 each reactant requires a different collision geometry that can be simple or complex
2. TEMPERATURE :
 an increase in temperature increases the rate of reaction
 as temperature rises, the reactant particles gain kinetic energy, moving faster,
colliding more frequently, and thus reacting more quickly
15

with a higher temperature, a larger fraction of the molecules will have the required
kinetic energy to have effective collisions
3. CONCENTRATION:
 an increase in reactant concentration increases the rate of reaction
 the greater the concentration, the greater the number of particles per unit volume,
which are more likely to collide as they move randomly within a fixed space
4. SURFACE AREA:
 an increase in reactant surface area increases the rate of reaction
 reactants can collide only at the surface where the substances are in contact, and
by increasing the surface area, you are increasing the number of particles in an
area, thereby increasing the probability of an effective collision
5. CATALYST:
 a catalyst is a substance that increases the rate of a chemical reaction without itself
being permanently changed
 a catalyst provides an alternate “pathway”, with lower activation energy, to the same
product formation, meaning a much larger fraction of collisions are effective
 the catalyst can help break the bonds in the reactant particles, provide a surface for
the necessary collisions, and allow the reactants’ atoms to recombine in new ways
 catalysts are involved in the reaction mechanism at some point, but are regenerated
before the reaction is complete
CHAPTER 7: CHEMICAL SYSTEMS IN EQUILIBRIUM
equilibrium – the balanced state of a reversible reaction or process where there is no
net observable change; the rate of the forward reaction equals that of the
reverse reaction (A ↔ B)
– can be approached from either side of the reaction equation
– the concentration of the reactants and products do not change (are
constant)
solubility equilibrium – an equilibrium between a solute and a solvent in a saturated
solution
phase equilibrium – an equilibrium between different physical states of a pure
substance (e.g., ice over a lake)
chemical reaction equilibrium – an equilibrium between reactants and products of a
chemical reaction
EQUILIBRIUM LAW EQUATION
For the general chemical reaction:
K = [C]c [D]d
(liquids and
[A]a [B]b
aA + bB ↔ cC + dD
where: • A, B, C, D are chemical entities in gas or aqueous phases
solids are omitted from the equation)
• a, b, c, and d are the coefficients in the balanced chemical
equation
• K is the equilibrium constant (temperature and pressure
specific)
- if K is large, reaction’s concentration of products greater than
reactants
16
- if K is small, reaction’s concentration of reactants greater than
products
- K is inversely proportional to the K value of the reverse reaction
LE CHÂTELIER’S PRINCIPLE
Le Châtelier’s Principle – when a chemical system at equilibrium is disturbed by a
change in a property, the system adjusts in a way that
opposes the change
equilibrium shift – movement of a system at equilibrium resulting in a change in the
concentrations of reactants and products
VARIABLES AFFECTING CHEMICAL EQUILIBRIA:
VARIABLE
TYPE OF CHANGE
concentration
increase
decrease
temperature
increase
RESPONSE OF SYSTEM
shifts to consume added reactant
shifts to replace removed reactant
shifts to consume added thermal energy
(away from heat term)
shifts to replace removed thermal
energy (towards heat term)
shifts towards side with larger total
number of gaseous entities
shifts towards side with smaller total
number of gaseous entities
shifts away from common ion to
consume the added reactant
decrease
volume
increase (decrease in
pressure)
decrease (increase in
pressure)
common ion
dissolving a compound
effect
into solution that adds a
common ion
VARIABLES THAT DO NOT AFFECT CHEMICAL EQUILIBRIA
catalysts
no effect
adding inert
no effect
gases
SOLVING EQUILIBRIUM PROBLEMS
1.
2.
3.
4.
5.
6.
7.
8.
9.
Write a balanced equation for the reaction and list the known values.
If the direction the system must go to attain equilibrium is not obvious (i.e., one
entity is not present initially), calculate Q with the initial concentrations and compare
it to the value of K to determine which direction the system will proceed to attain
equilibrium.
Construct an ICE (Initial concentration, Change in concentration, Equilibrium
concentration) table and input the initial concentrations.
Let x represent the changes in concentration, multiplying it by the coefficient in the
balanced equation. The reactants should all change in the same way and all the
products should proceed in the opposite way.
Rewrite the E row using the x values.
Substitute equilibrium concentrations into the equilibrium constant equation.
Apply appropriate simplifying assumptions, if possible (e.g., 4x3 ÷ (0.4 – 2x)2 can be
simplified to 4x3 ÷ (0.4)2 because x value is so small in comparison).
Solve for x.
Justify any assumptions you have made (i.e., the x value you get should be plugged
into the original equation and the difference between the two must be less than
5%).
17
10. Calculate the equilibrium concentrations by substituting x into equilibrium
concentration expressions from the E row.
e.g., 4.00 mol of hydrogen and 2.00 mol of iodine are placed in a 2.00-L reaction vessel at
440°C and react to form hydrogen iodide. At this temperature, the K is 49.7. Determine
the concentrations of all entities.
[H2(g)] = 4.00 mol
2.00 L
= 2.00 mol/L
H2(g) + I2(g) ↔ 2 HI(g)
K = 49.7
[I2(g)] = 2.00 mol
2.00 L
= 1.00 mol/L
[HI(g)] = 0.00 mol/L
H2(g)
+
I2(g)
↔
2
HI(g)
Initial concentration (mol/L)
2.00
1.00
–x
–x
2.00 – x
1.00 – x
0.00
Change in concentration (mol/L)
+2
x
Equilibrium concentration
(mol/L)
K
=
49.7 =
2x
[HI]2
[H2] [I2]
[H2(g)] = 2.00 mol/L - x
= 2.00 mol/L - (0.93 mol/L)
= 1.07 mol/L
(2x)2
(2.00 – x) (1.00 – x)
0.92x2 – 3.00x + 2.00 = 0
[I2(g)] = 1.00 mol/L - x
= 1.00 mol/L - (0.93 mol/L)
= 0.07 mol/L
x = –b ± √ b2 – 4ac
2a
[HI(g)] = 2x
= 2(0.93 mol/L)
= 1.87 mol/L
4x = 49.7 (2.00 – x) (1.00 – x)
2
= 3.00 ± √ 9.00 – 7.36
1.84
= 2.33 or 0.93
2.33 is rejected, because concentrations cannot have a negative value (i.e., 2.00 – 2.33 = 0.33)
SOLUBILITY PRODUCT CONSTANT
solubility – the concentration of a saturated solution of a solute in a particular solvent
at a particular temperature; specific maximum concentration
solubility product constant (Ksp) – the value obtained from the equilibrium law applied
to saturated solution (remember solids are not
included in the equation); omit units as with all K
values
– can only be determined for ionic compounds that are
classified as insoluble or slightly soluble
1.
2.
3.
4.
Write a balanced equation and list the known values.
Use the solid product to write an equilibrium equation for it dissolving into ions.
Find the Ksp value for the solid product and write it next to the equilibrium equation.
Determine the number of moles of both ions, by using the mole ratios and initial
concentrations of reactants.
18
5. Determine the concentration upon mixing, by dividing the number of moles by the
new volume.
6. Plug these new concentrations into the Ksp equation to determine the Q
(experimental value).
7. Compare the Q value to the Ksp to predict whether a precipitate will form (see below).
USING Q TO PREDICT SOLUBILITY
Ion product, Q > Ksp
precipitate will form (supersaturated solution)
Ion product, Q = Ksp
precipitate will not form (saturated solution)
Ion product, Q < Ksp
precipitate will not form (unsaturated solution)
e.g., 20.0 mL of 0.20 mol/L ammonium sulfate solution is added to 130 mL of 0.50 mol/L
barium nitrate solution. What are the concentrations of the ions and will a precipitate
form?
(NH4)2SO4(aq) + Ba(NO3)2(aq)  BaSO4(s) + 2 NH4NO3(aq)
Choose the solid to write the
equilibrium equation (remember any
-10 NO3 molecule is aqueous)
BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)
Ksp = 1.08 x 10
nBa2NO3 = c v
= (0.50 mol/L) (0.130 L)
= 0.065 mol
= nBa
nNH4NO3 = c v
= (0.20 mol/L) (0.020 L)
= 0.0040 mol
= nSO4
[Ba2+(aq)] = n
v
Divide by the
=
0.065
mol
“new” volume (i.e.,
0.150
L
20 mL and 130 mL
=
0.43
mol/L
equals 150 mL)
Ksp = [Ba2+] [SO42-]
= (0.43) (0.027)
= 0.011
[SO42-(aq)] = n
v
= 0.004 mol
0.150 L
= 0.027 mol/L
0.011 > 1.08 x 10-10, therefore it will precipitate
ENTROPY
spontaneous reaction – one that, given the necessary activation energy, proceeds
without continuous outside assistance
entropy, S – a measure of the randomness or disorder of a system or the surroundings
– equals 0 when the temperature is at absolute zero (0 K)
FACTORS THAT INCREASE ENTROPY (S)
 the volume of a gaseous system increases (i.e., pressure decreases)
 the temperature of a system increases
 the physical state of a system changes from solid to liquid to gas, or liquid to gas
(i.e., Sgas > Sliquid > Ssolid)
 fewer moles of reactant molecules form a greater number of moles of product
molecules
 complex molecules are broken down into simpler subunits (e.g., combustion of
organic fuels into carbon dioxide and water)
CLASSIFICATION OF SPONTANEOUS AND NONSPONTANEOUS REACTIONS
Entropy increases (ΔS > 0)
Endothermic (ΔH > 0)
spontaneous at high temps.
nonspontaneous at low
19
Exothermic (ΔH < 0)
spontaneous
C(s) + O2(g)  CO2(g)
temps.
H2O(s)  H2O(l)
Entropy decreases (ΔS < 0)
spontaneous at low temps.
nonspontaneous at high
temps.
nonspontaneous
3 O2(g)  2 O3(g)
2 SO2(g) + O2(g)  2 SO3(g)
Gibb’s free energy, G – energy that is available to do useful work; ΔG° = ΔH° – (T
ΔS°)
CHAPTER 8: ACID-BASE EQUILIBRIUM
BRØNSTED-LOWRY THEORY
Brønsted-Lowry acid – proton donor
Brønsted-Lowry base – proton acceptor
amphoteric (amphiprotic) – a substance capable of acting as an acid or a base in
different chemical reactions; a substance that may accept or
donate a proton conjugate pair
e.g.,
acid
HC2H3O2(aq) + H2O(l) ↔ C2H3O2-(aq) + H3O+(aq)
base
strong acid – an acid with a very weak attraction for protons and easily donates it to a
base
strong base – a base with a very strong attraction for protons

The stronger an acid, the weaker its conjugate base, and vice versa.
AUTOIONIZATION OF WATER
autoionization of water – the reaction between two water molecules producing a
hydronium ion and a hydroxide ion (H2O(l) ↔ H+(aq) + OH-(aq)
Kw = [H+(aq)] [OH-(aq)]
[H+(aq)] =
= (1.0 x 10-7 mol/L) (1.0 x 10-7 mol/L)
[H+(aq)]
= 1.0 x 10-14
Kw
[OH-(aq)] =
Kw
[OH-(aq)]
In neutral solutions [H+(aq)] = [OH-(aq)]
In acidic solutions [H+(aq)] > [OH-(aq)]
In basic solutions [H+(aq)] < [OH-(aq)]
PH
pH – the negative of the logarithm to the base ten of the concentration of hydrogen
(hydronium) ions in a solution
pOH – the negative of the logarithm to the base ten of the concentration of hydroxide
ions in a solution
pH = –log [H+(aq)]
[H+(aq)] = 10–pH
pOH = –log [OH-(aq)]
[OH-(aq)] = 10–pOH
20
pH + pOH = 14.00
e.g.,
Calculate the pH of a solution prepared by dissolving 4.3 g of Ba(OH)2(s) in water to form
1.5 L of solution.
Ba(OH)2(aq)
100%
Ba2+(aq) + 2 OH–(aq)
nBa(OH)2 = 4.3 g x 1 mol
171.3 g
= 2.5 x 10-2 mol
[Ba(OH)2(aq)] = 2.5 x 10-2 mol
1.5 L
= 1.7 x 10-2 mol/L
[OH-(aq)] = 2 (1.7 x 10-2 mol/L)
= 3.3 x 10-2 mol/L
pOH = –log [OH-(aq)]
= –log (3.3 x 10-2)
= 1.47
pH = 14.00 – pOH
= 14.00 – 1.47
= 12.53
STRONG VS. WEAK ACIDS AND BASES
strong acid – an acid that is assumed to ionize quantitatively (completely) in aqueous
solution (i.e., percent ionization is > 99%); HCl(aq), HNO3(aq), and H2SO4(aq)
are the only strong acids we will work with
strong base – an ionic substance that dissociates completely in water to release
hydroxide ions; all of the metal hydroxides are strong bases
weak acid – an acid that partially ionizes in solution but exists primarily in the form of
molecules
weak base – a base that has a weak attraction for protons
percent ionization (p) = concentration of acid ionized x 100%
[HA(aq)]
concentration of acid solute
[H+(aq)] =
p
100
acid ionization constant (Ka) – equilibrium constant for the ionization of an acid
e.g., Calculate the Ka and pH of hydrofluoric acid if a 0.100 mol/L solution at equilibrium at
SATP has a percent ionization of 7.8%.
HF(aq)
7.8%
H+(aq) + F-(aq)
Ka = [H+(aq)] [F-(aq)]
[HF(aq)]
[H+(aq)] = (p/100) [HA(aq)]
= (7.8 / 100) (0.100 mol/L)
= 0.0078 mol/L
↔
HF(aq)
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration
0.100
0.000
–x
0.100 – 0.0078
21
H+(aq)
+
F-(aq)
0.000
+x
0.0078
+x
x
(mol/L)
Ka = [H+(aq)] [F-(aq)]
[HF(aq)]
= 0.00782
0.0922
= 6.6 x 10-4
0.0078
= 0.0922
pH = –log [H+(aq)]
= –log (0.0078)
= 2.1
CHAPTER 9: ELECTROCHEMISTRY (ELECTRIC CELLS)
OXIDATION NUMBER
oxidation number – an integer that is assigned to each atom in a compound when
considering redox reactions
– a positive or negative number corresponding to the apparent charge
that an
atom in a molecule or ion would have if the electron pairs in covalent
bonds belonged entirely to the more electronegative atom
RULES FOR ASSIGNING OXIDATION NUMBERS
1. The oxidation number of an atom in an uncombined element is always 0 (e.g., H2 is
0).
2. The oxidation number of a simple ion is the charge of ion (e.g., Ca2+ is +2).
3. The oxidation number of hydrogen is +1, except in metal hydrides when it is -1
(e.g., the H in NaH is -1).
4. The oxidation number of oxygen is -2, except in peroxides when it is -1 (e.g., the O
in H2O2 is -1).
5. The oxidation number of Group 1 element ions is +1. The oxidation number of
Group 2 element ions is +2.
6. The sum of oxidation numbers in a compound must equal 0.
7. The sum of oxidation numbers in a polyatomic ion must equal the charge on the ion
(e.g., OH- is -1).
OXIDATION-REDUCTION
Oxidation can be defined as a reaction in which:
1) an element is chemically united with oxygen (e.g., C + O2  CO2; carbon is
oxidized)
2) a metal is changed from an uncombined to a combined state (e.g., Zn + Cl2 
ZnCl2)
3) an element loses electrons, and therefore has an increase in oxidation number
(e.g., Ca  Ca2+ + 2e-)
Loss of Electrons is Oxidation “LEO”
Reduction can be defined as a reaction in which:
1) an element loses oxygen (e.g., Fe2O3  2 FeO + ½ O2)
2) a metal is changed from a combined to a uncombined state (e.g., FeO  Fe +
½ O2)
1) an element gains electrons, and therefore has a decrease in oxidation number
(e.g., Cl2 + 2e-  2 Cl-)
Gain of Electrons is Reduction “GER”
22
redox reaction – a chemical reaction in which electrons are transferred between
particles; two or more atoms undergo a change in oxidation number;
also known as oxidation-reduction reactions
– all single displacement reactions are redox, while some combination
and
decomposition reactions are; double displacement reactions are never
redox
e.g.,
oxidation
+1 -2
H2S(g)
0
+
+4 -2

O2(g)
+1 -2
SO2(g)
+
H2O(g)
reduction
BALANCING REDOX EQUATIONS USING OXIDATION NUMBERS

this method is most appropriate when dealing with covalent compounds
1. Assign oxidation numbers to all the atoms in the equation.
2. Identify which atoms undergo a change in oxidation number.
3. Determine the ratio in which these atoms must react so that the total increase in
oxidation number equals the decrease (i.e., the total number of electrons lost and
gained is equal).
4. Balance the redox participants in the equation using this ratio.
5. Balance the other atoms by the inspection method.
6. Add H+(aq) or OH-(aq) to balance the charge, depending on if it is an acidic or a basic
solution (the total charge on each side must be the same).
7. Add H2O(l) to balance the O atoms.
e.g.,
Ag(s) + Cr2O72-(aq)  Ag+(aq) + Cr3+(aq)
oxidation: lost 1 e- (x 6)
0
6 Ag(s)
+
+6 -2
Cr2O72-(aq)
+ 14 H+(aq)

+1
6 Ag+(aq)
+
+3
2 Cr3+(aq)
+
H2O(l)
reduction: gained 2(3 e-) = 6 e- (x 1)
BALANCING REDOX EQUATIONS USING HALF-REACTIONS

this method is most appropriate for ionic reactions in solution and for relating to
electrical processes
1. Separate the skeleton equation into the start of two half-reaction equations (one for
the oxidation reaction and one for the reduction reaction).
2. Balance all species, other than O and H.
3. Balance the oxygen, by adding H2O(l) for acidic solutions or OH-(aq) for basic
solutions.
4. Balance the hydrogen, by adding H+(aq) for acidic solutions or H2O(l) for basic
solutions.
5. Balance the charge on each side by adding electrons and canceling anything that is
in equal amounts on both sides.
23
6. Multiply each half-reaction equation by simple whole numbers to balance the
electrons lost and gained.
7. Add the two half-reaction equations, canceling the electrons and anything else that
appears in equal amounts on both sides.
8. Check to ensure all entities and the overall charge on both sides balance.
MnO4– + N2H4  MnO2 + N2
e.g.,
4 [MnO4– + 2 H2O + 3 e-  MnO2
3 [N2H4 + 4 OH–  N2
+
4 H2O
4 MnO4– + 8 H2O + 12 e-  4 MnO2
3 N2H4 + 12 OH–  3 N2
+
12 H2O
+ 4 OH–]
+ 4 e-]
+ 16 OH–]
+ 12 e-]
4 MnO4– + 3 N2H4  4 MnO2 + 3 N2 + 4 H2O + 4 OH–
TECHNOLOGY OF CELLS AND BATTERIES
electric cell – a device that continuously converts chemical energy into electrical
energy; contains two electrodes (solid conductor) and one electrolyte
(aqueous conductor); each electrode has a cathode (+) and anode (–);
electrons flow from the anode to the cathode
battery – a group of two or more electric cells connected in series
voltage – the potential energy difference per unit charge; measured in volts (V), 1 J/C
electric current – the rate of flow of charge past a point; measured in amperes (A), 1
C/s
TYPE
primary
cell –
electric cell
that cannot
be
recharged
secondary
cell –
electric cell
that can be
recharged
NAME OF
CELL
HALF-REACTIONS
dry cell
(1.5 V)
2 MnO2 + 2 NH4+ + 2 e-  Mn2O3 + 2 NH3 +
H2O
Zn  Zn2+ + 2 e-
alkaline dry
cell (1.5 V)
2 MnO2 + H2O + 2 e-  Mn2O3 + 2 OHZn + 2 OH-  ZnO + H2O + 2 e-
mercury cell
(1.35 V)
HgO + H2O + 2 e-  Hg + 2 OHZn + 2 OH-  ZnO + H2O + 2 e-
Ni-Cad cell
(1.25 V)
2 NiO(OH) + 2 H2O + 2 e-  2 Ni(OH)2 + 2
OHCd + 2 OH  Cd(OH)2 + 2 e-
lead-acid cell
(2.0 V)
PbO2 + 4 H+ + SO42- + 2 e-  PbSO4 + 2 H2O
Pb + SO42-  PbSO4 + 2 e-
24
CHARACTERISTICS
AND USES
 inexpensive,
portable
 flashlights, radios
 longer shelf life,
higher currents for
longer periods
 same uses as dry
cell
 small cell, constant
voltage during life
 hearing aids,
watches
 completely sealed,
lightweight
 power tools,
shavers, portable
computers
 large currents,
reliable for
recharges
 all vehicles
fuel cell –
electric cell
that
produces
electricity
by a
continually
supplied
fuel
aluminum-air
cell (2 V)
3 O2 + 6 H2O + 12 e  12 OH
4 Al  4 Al3+ + 12 e-
hydrogenoxygen cell
(1.2 V)
O2 + 2 H2O + 4 e-  4 OH2 H2 + 4 OH-  4 H2O + 4 e-
-
-
 high energy
density, readily
available aluminum
alloys
 electric cars
 lightweight, high
efficiency, can be
adapted to use
hydrogen-rich fuels
 vehicles and space
shuttle
GALVANIC CELLS
galvanic cell – consists of two halfcells separated by a
porous boundary with
solid electrodes
connected by an
external circuit;
standard cells occur at
SATP and with
concentrations of 1.0
mol/L
cathode – positive electrode;
reduction (gaining electrons
because electronegativity is
higher) of the strongest
oxidizing agent occurs here;
the half-cell that is higher
on the “Relative Strengths
of Oxidizing and Reducing
Agents” table; “red cat on
the roof”  reduction at the cathode, higher half-cell
anode – negative electrode; oxidation (losing electrons because electronegativity is
lower) of the strongest reducing agent occurs here; the half-cell that is lower on
the table
inert electrode – a solid conductor that will not react with any substance present in a
cell (usually carbon or platinum)


Electrons travel in the external circuit from the anode to the cathode
Internally, anions from the salt bridge move toward the anode and cations from the
salt bridge move toward the cathode as the cell operates, keeping the solution
electrically neutral
CELL NOTATION
electrons
cathode (+) | electrolyte || electrolyte | anode (-)
(reduction)
(oxidation)
CELL POTENTIALS
25
standard cell potential (ΔE°) – the maximum electric potential difference (voltage) of a
cell operating under standard conditions
reference half-cell – a half-cell assigned an electrode potential of exactly 0.00 volts;
the standard hydrogen half-cell, Pt(s) | H2(g), H+(aq)
standard reduction potential (ΔEr°) – represents the tendency of a standard half-cell
to attract electrons in a reduction half-reaction,
compared to the reference half-cell
standard oxidation potential (ΔEo°) – represents the tendency of a standard half-cell
to lose electrons in an oxidation half-reaction; the
value of the reverse reaction with an opposite sign
ΔE° =
cell

ΔEr°
cathode
–
ΔEr°
anode
A positive standard cell potential (ΔE° > 0) indicates that the overall cell reaction is
spontaneous.
e.g.,
SOA
OA
Ag(s) | Ag+(aq) || Cu2+(aq) | Cu(s)
RA
SRA
2 [Ag+(aq) + e-  Ag(s) ]
Cu(s)  Cu2+(aq) + 2 eCu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
ΔE° =
ΔEr°
cell
ΔE° =
=
=
–
ΔEr°
cathode
anode
ΔE°Ag+ – ΔE°Cu2+
0.80
–
0.34
0.46 V
26